process 5. dynamics of simple systems - Åbo akademi | … · · 2014-02-055. dynamics of simple...

5. Dynamics of Simple Systems 5.1 Integrating systems

5.2 First-order systems 5.2.1 Transient response 5.2.2 Identification from step response

5.3 Second-order systems 5.3.1 Transient response 5.3.2 Identification of overdamped systems 5.3.3 Identification of underdamped systems

5.4 Time-delay systems

5.5 Inverse-response systems

5.6 Systems in series

KEH Process Dynamics and Control 5-1

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5. Dynamics of Simple Systems In Chapter 3, we derived models for some simple technical systems. In all cases, the systems could be described by ordinary differential equations (ODEs) of relatively low order

In many cases, the ODEs were nonlinear, but they could be linearized around an operating point

In this chapter we shall study the properties of some simple, linear, dynamical systems

In particular, we shall derive the time response to well-defined inputs such as impulses and steps;

analysis based on such responses is called transient analysis We shall also study simple, graphical techniques for determining linear low-order models

from step response data; these are simple examples of system identification


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5. Dynamics of Simple Systems 5.1 Integrating systems

An integrating system is the simplest dynamical system that can be described by a differential equation. The most typical process example of an integrating system is probably a liquid tank.

Example 5.1. A liquid Tank.

Consider the tank in figure 5.1.

V = volume of liquid in the tank

F1 = volumetric flow rate of liquid into the tank

F2 = volumetric flow rate of liquid out of the tank Figure 5.1. Liquid tank.

Notice that V is the output signal of the system (i.e., a dependent variable), whereas F1 and F2 are input signals (“independent” variables). See section 2.3.


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5.1 Integrating systems A mass balance around the tank gives under the assumption of constant density (which cancels out) the model

(1)

This equation is linear and we can directly replace the variables with Δ-variables to get (2)

The Laplace transform considering that the initial state is zero gives

or (3)

The two transfer functions of the system are

and (4)

which, according to the Laplace transform , corresponds to integrals in the time domain.


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1 2ddV F Ft= −

1 2d

dV F Ft∆

= ∆ − ∆

1 2( ) ( ) ( )s V s F s F s∆ = ∆ − ∆ 1 21 1( ) ( ) ( )V s F s F ss s

∆ = ∆ − ∆

1

( ) 1( )

V sF s s

∆=

∆ 2

( ) 1( )

V sF s s∆

= −∆

5.1 Integrating systems

Generally, a linear integrating system with input signal u and output signal y can be described by the differential equation

or (5.1)

The transfer function of the system is

(5.2)

Exercise 5.1.

Derive and sketch (a) the impulse response (b) the step response (c) the ramp response

for ΔV(t) when a change in the input flow F1 to the tank in Fig. 5.1 is done.


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Kuty=

dd

utyT =

dd

TssK

sUsYsG 1)()()( ===

5. Dynamics of Simple Systems 5.2 First-order systems

A linear first-order system can be described by differential equation

(5.3)

where K is the system gain and T its time constant. The transfer function of the system is

(5.4)

5.2.1 Transient response

The system time response y(t) of a given input signal u(t) can be easily determined by finding the inverse Laplace transform by using the Table of Laplace transforms (see section 4.2). Two common inputs are the

impulse function

step function


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KuytyT =+

dd

11

)()()(

+==

TssUsYsG


Impulse response

If the input signal of the system is an impulse with the time integral (“area”) I, i.e. u(t) = I δ(t)

where δ(t) is the unit impulse (Dirac delta function), its Laplace transform (as listed in the table of Laplace transforms) is

U(s) = I

The inverse Laplace transform of

then gives the impulse response

(5.5)


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1)()()(

+==

TsKIsUsGsY

t/T

TKIty −= e)(


Step response

If the input signal is a step change of magnitude ustep , i.e.

u(t) = ustepσ(t)

where σ(t) is the unit step, the Laplace transform is

The inverse Laplace transform of

then gives the step response

(5.6)


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step( )u

U ss

=

step( ) ( ) ( )( 1)Ku

Y s G s U sTs s

= =+

/step( ) (1 e )t Ty t Ku −= −



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Figure 5.2. Impulse response of a first-order system.

Figure 5.3. Step response of a first-order system.

− The initial slope of the curves (i.e. their derivative) is obtained by drawing a line from the start of the response so that it intersects the line representing the new steady-state value at t = T .

− The distance between the response and the point of intersection at t = T is 1/e = 0.368 of the total change of the output signal.

− In practice, the new steady-state value (within 2%) is reached at t ≈ 4T (however, in theory it takes ∞ long time).

5.2 First-order systems 5.2.2 Identification from step response

From the information above it is evident that the system gain and time constant can be identified (i.e. determined) from the transient response, which can be generated experimentally by a suitable change in the input signal.

In this respect, the identification from step changes are normally used, particularly because a well defined impulse response is difficult to achieve.

The identification by a step input gives the system gain according

K = y∞ / ustep (5.7)

• ustep is the magnitude of the step change of the input signal

• y∞ is the total change of the output signal when t → ∞.

There are different methods for the determination of the system time constant and an eventual time-delay (see section 5.4). Some simple “graphical” methods will be studied in the following section.


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5.2.2 Identification from step response 63 % of the total change

− The time constant of a first order system can be determined starting from the point of intersection between the final asymptotic value (y∞ = Kustep) and the tangent (i.e. the derivative) to that point where the step response begins to change (see figure 5.3).

− However, it is difficult in practice to determine the tangent (i.e. the step response initial derivative) with good accuracy.

It is better to use the point where the step response reaches 63,2 % of the total change.

− One can easily show that the step response of a first order system reaches this point when a time equal to the system time constant has elapsed since the beginning of the step response. In other words, the time constant is given by the time coordinate of that point where 63,2 % of the total change has been reached.

− Generally, the time constant that is obtained from the 63,2 % of the total change can be called equivalent time constant even if the system is not of first order.


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5.2.2 Identification from step response In practice, a system often contains a time delay, for example due to a transport delay. The step response delays then with the corresponding time, which should be considered when identifying the system .


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A first order system with a time delay L has the transfer function (5.8) and the step response (5.9) A step change at t = 0 gives a step response that starts at time L and reaches 63,2 % of the total change at time L + T . Both parameters are thus obtained from the same step response. In figure 5.4 the step response has been normalized by dividing the output signal by y∞ .

1e

)()()(

+==

−

TsK

sUsYsG

Ls

)e1()( /step

TtKuLty −−=+

Figure 5.4. Identification of a 1st order system from the 63 % of the total change.

5.2.2 Identification from step response The Tangent method

In practice, a perfect first-order system is hardly ever obtained (with or without time delay).

− Often the step response does not have its steepest slope immediately at the beginning, which a system of first order would have.

− It means that the system is of higher order than first order.


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Sometimes we want anyways to approximate the system to a first-order system with time delay. Figure 5.5 illustrates this method.

− The system gain is calculated in the normal way according to equation (5.7).

− The time delay and time constant are obtained by drawing a tangent through the inflection point of the step response (ti ,yi), i.e. the steepest part of a slope. The tangent point of intersection with the time-axis gives the time delay L , the intersection point with the final asymptotic value gives the time coordinate L + T.

Figure 5.5. Identification of a 1st order system with the tangent method.

5.2.2 Identification from step response − Ziegler-Nichols’ as well as some other step-response based

recommendations for setting of PID controllers (see section 7.5) assume that the model parameters are determined according the tangent method.

− As shown in figure 5.5, the model step response (the dotted line) can, however, differ significantly from the actual step response. Because the step response of a first-order system has its steepest slope at the beginning, where it is the same as the slope of the drawn tangent, and then decreases, it is easy to understand that the model step response will always be below the line of the actual step response.

− In other words, the time constant determined by the tangent method is too large.

− However, this does not mean that the method is not ok for controller setting, but it is relatively poor for model identification.


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5.2.2 Identification from step response Modification of the tangent method

A considerable modification of the methods named above can be obtained if both of them are combined. We determine then • the time delay according the

tangent method • the time constant = the equivalent

time constant, i.e. the time it takes for the step response to reach 63,2 % of the total change.


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− This procedure gives a model whose step response ( the dotted line in figure 5.6) matches much better the actual step response.

− This method is also less sensible to disturbances, because two points of the step response are used to determine the model parameters.

− The ordinary tangent method try to determine both time delay and time constant from the step response characteristics in only one point, the inflection point, which is also difficult to handle in practice.

Figure 5.6. Identification of a 1st order system with the modified tangent method.

5.2.2 Identification from step response Sundaresan-Krishnaswamy method

− The inflection point and the point where the step response reaches 63,2 % of the total change are often close to each other.

− A better adaptation can be expected if we use two points that are slightly further apart.

− According Sundaresan and Krishnaswamy (1977) we will use the two points where the actual step response reaches 35% and 85% of the total change.


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If the time coordinate for each point is denoted t35 and t85 , we can then derive by using equation (5.9) T = 0,682(t85 - t35) (5.10) L = t35 – 0,431T (5.11) The gain K is calculated according equation (5.7)

Figure 5.7. Identification of a 1st order system from 35% and 85% of the change.

5.2.2 Identification from step response The Logarithmic method

There is a simple way to check how well an experimental step response is consistent with the step response of a first-order system (with or without time delay) without actually determining the model parameters.

We can derive the following relationship from equation (5.9)

, y∞ = Kustep (5.12)

− If the expression to the left of the equation plots as function of t , we get then for a first-order system a straight line that has slope coefficient -1/T and intersects the time axis (i.e. the value is 0 ) in the point t = L.

Same expression can be calculated and plotted for an arbitrary experimental step response.

− If the obtained relationship is sufficiently linear the system is then of first order. • At the same time, the system time constant is obtained based on the

slope coefficient of the straight line, and its possible time delay from the intersection point of the line with the time axis.


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TLt

ytyy −

−=

−

∞

∞ )(ln

5.2.2 Identification from step response − If the relationship is moderately nonlinear we might consider to determine

an approximate model of first order by fitting the relationship with a straight line.

• One is tempted to draw a line so that the final asymptotic value coincides with the plotted experimental relationship when t tends to infinity.

• However, this gives a too large time delay and too small time constant. A smaller slope must be chosen.


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Figure 5.8. Identification of a 1st order system with the logarithmic method.

Figure 5.9. Step response of a 1st order system identified with the logarithmic method.

5.2.2 Identification from step response In brief, we can say that

• the modified tangent method

• the method proposed by Sundaresan and Krishnaswamy (the points 35% and 85%) are clearly the bests of the simple graphics methods here presented for identification of a first-order system with time delay.


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5. Dynamics of Simple Systems 5.3 Second-order systems

A strictly linear system of second order can be described with the differential equation

(5.13)

and the transfer function

(5.14)

We will only consider systems with b2 ≠ 0 and in this section only the case where b1 = 0, i.e. systems with transfer functions without zeros.

In section 5.5 we will consider systems with b1 ≠ 0.


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ubtubya

tya

ty

21212

2

dd

dd

dd

+=++

212

21

)()()(

asasbsb

sUsYsG

+++

==

5.3 Second-order systems To emphasize the overall properties of the system the transfer function is often written in the form

(5.15)

where K is the system gain, ζ is called relative damping and ωn is a constant called undamped natural frequency. Sometimes, the following form is also used

(5.14)

where Tn = 1/ ωn . There is no universally accepted term for Tn , both natural period and second-order time constant are used.

The system is said to be underdamped if 0 ≤ ζ < 1 , critically damped if ζ = 1 and overdamped if ζ > 1. If ζ < 0 the system is unstable.

KEH Process Dynamics and Control 21

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2nn

2

2n

2)(

ωζωω

++=

ssKsG

12)(

n22

n ++=

sTsTKsGζ

5.3 Second-order systems 5.3.1 Transient response

The transient response to a given input signal change can be determined in a normal way by using the Laplace transform inversion.

Here should be considered that the form of the solution is different depending on if the system is

• underdamped

• overdamped

• critically damped

The reason is that

• an underdamped system (0 ≤ ζ < 1) has complex poles (i.e. roots of the characteristic equation are complex)

• an overdamped system (ζ > 1) has real poles

• a critically damped system (ζ = 1) has a real double pole


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5.3.1 Transient response Critically damped systems

The transfer function of a critically damped system is often written in the form

(5.17)

where Tn = 1/ωn . The impulse and step response are obtained by using the Laplace transform inversion of the expression Y(s) = G(s)U(s).

For an impulse of magnitude I , is then U(s) = I , which gives the impulse response

(5.18)

For a step change of magnitude ustep corresponds U(s) = ustep / s , which gives the step response

y(t) = Kustep (1 - (1 + t / Tn) e-t / Tn) (5.19)

The responses are represented in figure 5.10 and 5.11 (ζ = 1) .


2n )1(

)(+

=sTKsG

n/2

n

)( TteT

tIKty −=

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5.3.1 Transient response Overdamped systems

The transfer function of a overdamped system is written often in the form

(5.20)

where T1 and T2 are related to ωn and ζ according (assuming T1 >T2 )

(5.21) (5.22)

The system has the impulse response

(5.23)

and the step response

(5.24)

The responses are represented in figure 5.10 and 5.11 (ζ > 1) . KEH Process Dynamics and Control 5-24

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)1)(1()(

21 ++=

sTsTKsG

1

, 1

n

2

2n

2

1 ωζζ

ωζζ −−

=−+

= TT 2

, 1

21

21

21 TTTT

TTn+

== ζω

( ) ee )( 21 --

21

t / Tt / T

TTIKty −

−=

( ) ee11)( 21 -2

-1

21step

−

−−= t / Tt / T TT

TTKuty

5.3.1 Transient response Underdamped systems

The fact that the characteristic equation of an underdamped system has complex roots makes the analytical expression of the system transient response to contain trigonometric functions.

The impulse response is given by

y(t) = K I ωn β-1 e-ζ ωnt sin(β ωnt) (5.25)

where

(5.26)

The step response is given by

y(t) = Kustep (1 - β-1 e-ζ ωnt sin(β ωnt + ϕ) (5.25)

Alternatively, the step response can be expressed by using trigonometric relationship as sin(βωnt) and cos(βωnt) (or another form of the Laplace transform) . The responses are represented in figure 5.10 and 5.11 (ζ < 1) .


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10 , 1 2 <≤−= ζζβ

5.3.1 Transient response Normalized transient response

Figure 5.10 shows the impulse response and figure 5.11 the step response of different systems of second order without zeros. When the response and the time are normalized, as shown in the figures, the responses are clearly determined by the damping factor ζ.

The transient responses of an underdamped system oscillate, while the responses of a critically damped system and an overdamped system are monotone.


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Figure 5.11. Step response of a system of second order without zeros.

Figure 5.10. Impulse response of a system of second order without zeros.

5.3 Second-order systems 5.3.2 Identification of overdamped systems

A simple method is described here for identification of a overdamped second order system without zeros by using its step response. The system transfer function is given by equation (5.20), or if a time delay is include (5.29) − The system gain K is determined as previously according equation (5.7). − Any time delay is given by the time that the initial value of the step

response is delayed in relation to the step change. − The main problem is thus to determine the system time constants T1 and

T2 . We will consider the case T1 ≥ T2 . As a limiting case, the system can be critically damped (T1 = T2 ), of first order (T2 = 0).


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)1)(1(e )(

21 ++=

−

sTsTKsG

Ls

5.3.2 Identification of overdamped systems

Modified Harriott’s method

Harriott (1964) has developed a relatively simple graphic method for determining the transfer functions of the type (5.29) by using the step response. Since numerical calculations of the type that form the basis of this method is not a problem nowadays, we will introduce here a slightly improved version of the Harriott’s method.

General description − All systems of the type (5.29) have a step response that reaches 72 % of the

final total change at a time t = L + 1,25(T1 + T2). If we first estimate the time delay L we can easily get the sum of the time constants from the step response at this point.

− The step responses of systems with different values of the parameter z = T1/(T1 + T2) are well separated at the time t = L + 0,5(T1 + T2). The parameter z provides a good characteristic of the system properties because

• A first order system has z = 1 , • A critically damped second order system has z = 0,5 , • for an overdamped second order system applies 0,5 < z < 1 .


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Figure 5.12 shows the step response of a first order system, a critically damped second order system and a overdamped second order system with z = 0,8 .


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− The step responses are normalized so that the output signal y is divided by the final change y∞ and the time is indicated by the variable τ = (t – L)/(T1 + T2) .

− The step responses reach 72% of the total change at τ = τ72 ≈ 1,25 and they are well separated at τ = τz ≈ 0,5 . Figure 5.12. Step response of a overdamped

second order system with different values of z.

− The sum of the time constants ∑Ti can be estimated by using the time τ72 , and the step response value at τz can be used for the estimation of the parameter z according to the graph in figure 5.13.

− When ∑Ti = T1 + T2 and z =Ti / ∑Ti are known the time constants T1 and T2 can then be calculated.


− It is thanks to the time axis normalization in figure 5.12 that the step responses reach 72 % at the same normalized time τ72 and they have a good separation at another normalized time τz .

− This normalization requires that we know the sum of the time constants and the possible time delay, which we do not know, however, when identifying the system.

Procedure

Fortunately, the procedure is transformed so that it can be used with the actual time variable t .

− Indicate the time when the step response reaches 72 % of the total change according to the actual time scale with t72 .

− Estimate the sum of the time constants ∑Ti according

∑Ti = 0,8(t72 – L) (5.30)

where L is the time delay, which is estimated separately. One can often (to start with) assume that L = 0 , unless, it is obvious from the step response that some other value would be better.


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− Let tz denote the time in the actual time scale which is τz = 0,4τ72 , where the step responses from different types of systems deviate most from each other. This time is given by


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tz = 0,4t72 + 0,6L (5.31)

− Denote the value of the output signal at tz with yz ; find it from the step response.

− Calculate the ratio yz / y∞ , where the final change y∞ , as yz , can be read from the step response.

− Read z from the diagram in figure 5.13 which corresponds to yz / y∞

− If yz / y∞ < 0,27 , the time delay L has to be increased

Figure 5.13. yz for different values of z .

− If yz / y∞ > 0,4 , the time delay L has to be decreased − If a new value for L was selected, calculate a new tz etc.

− Calculate the time constants of the system

T1 = z ∑Ti , T2 = ∑Ti - T1 (5.32)


Iterative improvement

The procedure described above generally gives sufficient accurate estimate of T1 and T2 . The method is based on the assumption that t72 = 1,25 , but t72 varies slightly with z as shown in figure 5.14.


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If t72 , according to figure 5.14, deviates from the value 1,25, the estimate of the time constants can be improved as follows:

− Read 𝜏72 corresponding to z from figure 5.14.

− Calculate the new estimate of ∑Ti according to

∑Ti = (t72 – L) / τ72 (5.33)

− Calculate new estimates of the time constants according (5.32).

Figure 5.14. τ72 for different values of z .

Notice that the estimate of z , according to figure 5.13, is not affected because yz does not change.


Example 5.2. Approximate identification of 1st and 2nd order systems.

By using the unit step response of a system described by the transfer function

(1)

we will determine

a) an approximate model of first order with time delay according to the modified tangent method;

b) an approximate model of second order with a possible time delay according to the modified Harriott’s method.

For comparison's sake, we will also determine models optimally adapted to first and second order and compare the different step responses of the models with the exact step response.

For simplicity sake, we will calculate dimensionless times (i.e. we will not use any unit)


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)12)(14)(16(1)(

+++=

ssssG


The input signal u is an unit step, i.e. U(s) = 1 / s . We have, then, the unit step response (2) This expression is not in our table of Laplace transforms, which means that we need to do partial fractions. We omit details and we find out that the inverse transform of the generic expression (3) gives the time function (4) In our case, with T1 = 6 , T2 = 4 and T3 = 2, we get the step response (5)

as it is plotted in figure 5.15.


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sssssUsGsY

)12)(14)(16(1)()()(

+++==

ssTsTsTsF

)1)(1)(1(1)(

321 +++=

332211 /

2313

2/

3212

2/

3121

2

e))((

e))((

e))((

1)( TtTtTt

TTTTT

TTTTT

TTTTT

tf −−−

−−−

−−−

−−−=

2/4/6/ e21e4e

291)( tttty −−− −+−=


For both, a) and b) case, the gain K is needed.

For an unit step we have ustep = 1 , and according to figure 5.15 we have y∞ = 1. Equation (5.7) gives then the gain K = 1 .


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Figure 5.15. Unit step response of the system G(s).


a) We will determine a model of first order with time delay according to the modified tangent method.

− We begin by drawing a tangent through the point where the step response has its steepest slope, and by reading the point where the tangent intersects the time axis. The intersection point has the time coordinate t ≈ 2,5 , which gives the time delay L ≈ 2,5.

− At 63 % of the total change y∞ is y = y63 = 0,63 y∞ = 0,63. This value is obtained at t ≈ 12,5 (too approximately), which means that L + T ≈ 12,5.

We have determined a system of first order with K =1 , T = 10 and L = 2,5 , i.e. a system with the transfer function (6) The unit step U(s) = 1 / s and the inverse transform of Y1(s) = G1(s)U(s) gives the unit step response (7) Figure 5.16 shows this step response together with the step response of the actual system.


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s

ssG 5,2

1 e110

1)( −

+=

10/1 e1)5,2( tty −−=+


We can even determine the model parameters numerically by minimizing the square sum of the difference between the model step response and the step response of the actual system. Such optimization gives K =1,02 , T = 9,05 and L = 3,83.


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Figure 5.16. Unit step response of G(s) (solid line) and G1(s) (dotted line).

Figure 5.17. Unit step response of G(s) (solid line) and the optimized adjustment (dotted line).


b) We will determine a model of second order with the modified Harriott’s method.

− We begin by determining the time when the system reaches 72 % of the total change. According to 5.15, we get t72 ≈ 15.

− Based on the step response, it looks like if a time delay would be needed L ≈ 1. However, we get a better adjustment by choosing a time delay that is slightly larger than the “actual”, as is clear from the a)-case. Let us choose L = 1,5 . According to equation (5.30), we get then ∑Ti ≈ 10,8 .

− The next step is to determine tz according to equation (5.31), which gives here tz ≈ 6,9 . At this time, according to the step response in figure 5.15, is y = yz ≈ 0,275 , and z ≈ 0,6 (too approximately) is given by y∞ = 1.

− From figure 5.14, we can read then a revised τ72 ≈ 1,264 , which gives, according to equation (5.33), a sum of the time constants ∑Ti ≈ 10,68.

− Equation (5.32) gives T1 = z ∑Ti ≈ 6,41 and T2 = ∑Ti - T1 ≈ 4,27.


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We have determined a system of second order with the transfer function (8) that has the unit step response (9) This step response is represented in figure 5.18 together with the actual system step response.

− According to the figure the adjustment appear to be very god.

The adjustment to the real step response, for optimizing the parameters of a 2nd order system with time delay, gives K = 1,002 , T1 = T2 = 5,35 and L = 1,38. Figure 5.19 shows the step response of this system and the step response of the actual system.

− The adjustment is only marginally better than the one that was obtained with the modified Harriott’s method.


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s

sssG 5,1

2 e)127,4)(141,6(

1)( −

++=

( )27,4/41,6/2 27,441,6

14,211)5,1( tt eety −− −−=+



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Figure 5.18. Unit step response of G(s) (solid line) and G2(s) (dotted line).

Figure 5.19. Unit step response of G(s) (solid line) and the optimized adjustment (dotted line).

5.3 Second-order systems 5.3.3 Identification of underdamped systems

As shown in figure 5.11, a step response of an underdamped system (0 ≤ ζ < 1) is characterized by oscillations.

Obviously, the amplitude and frequency of the oscillations are used for identification of a second order underdamped system.

Systems with oscillatory step response can be characterized by different parameters that can be deduced from the step response. A number of such parameters are indicated in figure 5.20.

To facilitate the definition of the parameters, we assume that

• the initial value of the output signal is zero (i.e. we use deviation variable)

• the final value of the step response is positive (i.e. an input signal change so that the output signal will increase).


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5.3.3 Identification of underdamped systems

y∞ Output signal final value (>0)

ymax Output signal largest value, i.e. first peak of the overshoot.

M Maximum overshoot, M = (ymax − y∞) / y∞ .

P Time period of the oscillations (specially the first period).

tr Rise time = time required for the output signal to pass y∞ . Typically, is defined as the time that takes to change from 10 % to 90 % of y∞ .


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Figure 5.19. Step response of a underdamped system.

tδ Settling time, is the time it takes until the output signal has entered and remained within (1 − δ)y∞ and (1 + δ)y∞ , i.e. (1 − δ)y∞ ≤ y(t) ≤ (1 + δ)y∞ . Normally, δ = 0,05 = 5% or δ = 0,02 = 2% are used.


Based on the analytical solution of the system step response, we can derive an expression that relates those parameters to the parameters in the system transfer function (5.34) By using the notation: (5.35) for the maximum overshoot we get: (5.36) for the time period: (5.37) for the rise time: (5.38) These expressions are exactly derived. For the settling time applies approximately to (5.39) KEH Process Dynamics and Control 5-43

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10 , 2

)( 2nn

2

2n

2 <≤++

= ζωζω

ωss

KsG

21 ζβ −=

βπζ /max e−

∞

∞ =−

=y

yyM

n

2βωπ

=P

n

)/arctan(βω

ζβπ −=rt

δζωδ

δ >−≈ Mt , )ln(

n


Identification It is easiest, and sufficient, to measure M and P . − The system relative damping ζ can be determined by the equations (5.35)

and (5.36). − The undamped natural frequency ωn is obtained from equation (5.37). − The rise time and eq. (5.38) can even be used instead of (5.36) and (5.37). The system gain K is determined according to equation (5.7). Generally, the step response of a substantially underdamped system is sensitive to disturbances, variations in the parameters and deviations from assumptions of an ideal system. • mostly affects the initial response of the system and thus the first

overshoot • better results if the identification is based on several oscillations

We denote the nth overshoot’s maximum value with ymax,n and the nth undershoot’s minimum value with ymin,n . Using equation (5.27), we can derive

(5.40) where represents the ratio between the (n+k)th and the nth relative overshoot (or undershoot).


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βζπ /2R

min,

min,

max,

max, e kk

n

kn

n

kn Myy

yyyyyy −

∞

+∞

∞

∞+ ==−−

=−−

kM R


Example 5.3. Identification of an underdamped second order system.

We will identify an underdamped second order system based on the step response in figure 5.20. In the figure The time axis in the figure goes from 0 to 20 seconds and the output signal axis from 0 to 2,5.

From the figure, we obtain

Equation (5.35) and (5.36) can be solved with respect to ζ , which gives

(1)

Numerically, we get ζ = 0,2485. The undamped natural frequency is given equation (5.37) ωn = 0,998.

The gain K can not be determined because the magnitude of the input signal step is not given.

The correct values are ζ = 0,25 and ωn = 0,1.


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and 5,625,375,9447,05,1

5,117,2max =−≈=−

≈−

=∞

∞ Py

yyM

)(ln)ln(

22 MM

+

−=

πζ

5. Dynamics of Simple Systems 5.4 Time-delay systems The output signal from a system consisting only of a time delay L looks exactly like the input signal, but delayed with L time units.

If the output signal is denoted y(t) and the input signal u(t) , the response for a pure time delay is thus

y(t + L) = u(t) (5.41)

In practice, a time delay depends often on a transport delay.

− A typical example is a conveyor belt. − Even with liquid flows and gas flows in a pipeline occurs time delays on the

properties of the flowing medium such as temperature and concentration. − Measuring instruments can sometimes lead to a time delay, for example

when analysing measuring samples.

The transfer function of a time delay of magnitude L is

G(s) = e-Ls (5.42) This function is basically simple, but as we know, time delays cause problems in control engineering.


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5.4 Time-delay systems − Systems with time delay belong to non-minimum phase systems (see

chapter 8).

− That is why time delays give analytical problems and computational problems, especially in combination with other system elements.

− The reason is that the transfers functions for other types of system elements are rational functions, while the transfer function of a time delay is an irrational function. Therefore, we have a reason to use rational approximations of (5.42).

Simple rational approximations can be derived from the Taylor series of e-Ls , i.e.

(5.43)

The first two terms give the simple, but relatively imprecise, approximation

e-Ls ≈ 1 − Ls (5.44)

The more terms there are, the better the approximation. But, in practice, the handling of the expression becomes more difficult when the degree of the polynomial increases.


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...!3)(

!2)(1e

32

+−+−=− LsLsLsLs

5.4 Time-delay systems Another possibility is to rewrite the series expansion

(5.45)

If only the first two terms in the denominator are considered, we get then the approximation

(5.46)

which means that the time delay L is approximated to a first order system with time constant L . If more terms are included, we get then approximations with systems of higher order.


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...!3)(

!2)(1

1e1e 32

++++==−

LsLsLsLs

Ls

LsLs

+≈−

11e

5.4 Time-delay systems We can combine the methods in different ways. One way is to rewrite

(5.47)

and the Taylor series expansions of the nominator and the denominator. If only the first two terms of the Taylor series expansions are considered

(5.48)

i.e. a proper, but not strictly proper, first order system that has quite special properties, as shown by the right side of the expression above


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Ls

LsLs

21

21

e

ee−

− =

LsLs

LsLs

211

21

211211

e+

+−=+

−≈−

5.4 Time-delay systems Padé approximations are another type of approximations that are derived under some optimized conditions.

The first-order Padé approximation is identical to (5.48) while the second-order Padé approximation is

(5.49)

Note that (5.49) is not obtained starting from the cut-off Taylor series.

Padé approximations −there are also approximations of higher order − are derived so that their frequency responses (see chapter 8) are similar to the frequency response of the time delay (both have the gain K = 1 at all frequencies), while the time response deviates more.


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2

2

)(121

211

)(121

211

eLsLs

LsLsLs

++

+−≈−

5.4 Time-delay systems The exponential function can be defined by using the limit

(5.50)

If we rewrite e−Ls = 1 / eLs , we then obtain the following approximation

(5.51)

i.e. a nth order system where we can choose the order.

The order n = 1 gives the same approximation as (5.46). Higher order gives of course a better approximation.


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n

n

x

nx

+=

∞→1lime

nLs

Lsn

+

≈−

11

1e

5. Dynamics of Simple Systems 5.5 Inverse-response systems

− Systems with inverse response show step responses whose direction changes one or several times at the beginning of the step response.

− This shall not be confused with the oscillations of an underdamped system, whose step response oscillates around the desired value where the output signal reaches a steady state condition.

− Systems with inverse response belong to the group of systems that are known as non-minimum phase systems (see chapter 8).

Systems with inverse response are not uncommon.

− A simple example is the quicksilver thermometer. The glass tube expands first when the ambient temperature increases, causing the column of quicksilver to fall. The quicksilver shortly begins to expand (density decreases) and the level change starts going to right direction.

− Another example with the same type of behavior is the liquid level in a boiler when the feed water supply increases.


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5.5 Inverse-response systems Systems with inverse response are difficult to control because sometimes they give out misleading information.

Such systems are characterized by a transfer function with (one or more) positive zeros, which are equivalent to negative time constants in the numerator.


+

+

+

+−

+−=

+

+

+

+

+−=

+

+

+

+

+=

13

12

11

15,2

15,1

13

12

11

15,2

15,1

13

12

11

15,2

15,1

2

1

0

sTsTsT

sTsT

G

sTsTsT

sTsT

G

sTsTsT

sTsT

G

Figure 5.21. Step responses with different number of negative time constants in the numerator.

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5.5 Inverse-response systems Such examples suggest that inverse-response systems occur when two subsystems are connected in parallel whose gains have different signs.

Exercise 5.2.

Two subsystems with transfer functions

are connected in parallel so that a system with transfer function G = G1 + G2 is obtained.

Assume that T1 > T2 > 0 and show that G is a non-minimum phase system if


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( ) ( )11 2

22

1

11 +

=+

=sTKG

sTKG and

12

1

2

1 >−>KK

TT

5. Dynamics of Simple Systems 5.6 Systems in series

In the analysis of series-connected systems it is important to know if the systems are interfering or non-interfering.

• in an interfering system one subsystem is affected by the subsequent subsystem in the series

• in an non-interfering system each subsystem is only affected by previous subsystems in the series.

For instance, if we connect in series two copies of the lowpass filter in example 3.1, they will then

− interfere because the subsequent circuit charges the previous one.

− On the other hand, if we provide the first filter with a gain on the output side, then they will not interfere with each other.


5.6 Systems in series A similar situation can be obtained if we connect in series two container for liquids, where the outflow occurs due to gravity (comp. with example 3.5).

− If the outflow from the first container flows freely into the second one, then no interference occurs.

− If the containers are connected so that the outflow from the first container flows through a tube to the bottom of the second container, then interference occurs due to the counter-pressure of the liquid level exerted on the inflow in the second container.

To summarize, we can conclude the following about systems in series:

− Non-interfering subsystems in series are easy to handle. Their transfer functions can be derived separately and then combined by multiplication as shown in section 4.3.3.

− Interfering subsystems are difficult to handle, since the individual characteristics of each subsystem are modified by the interference. In such cases we must model and consider the subsystems as a whole.


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